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The maximum height reached by the ball is 99.2 m
Explanation:
When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (
towards the ground). Therefore, we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
In this problem, we have:
u = 44.1 m/s is the initial vertical velocity of the ball
v = 0 is the final velocity when the ball reaches the maximum height
s is the maximum height
is the acceleration of gravity (downward, so negative)
Solving for s, we find the maximum height reached by the ball:

Learn more about free fall:
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Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²
2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads
3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²
4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J
5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²
6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²
7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s
8. s = Θ r = 266 rads * 0.33m = 88 m
Answer:
The acceleration is 0.
Explanation:
The formula for acceleration is:
a = (v-u)/t
If v and u are equal, thus it would be 0 when subtracted, and anything divided by 0 is 0.
Acceleration is the way the motion is changing.